The steric number is equal to the number of $\sigma$-bonds + the number of lone pairs of electrons on the central atom. It gives us the number of hybridised orbitals.

It is pretty straight-forward to calculate it, but the problem here is that one must always draw the Lewis Structure before one can actually get to calculating the steric number, and then the number and types of hybrid orbitals. Even that is quite simple for a smaller compound, even like XeF$_6$, but when it comes to complex hydrocarbons, it's a little difficult.

My question is that is there any well-known (or not so well-known, but working) shortcut to doing this, so as to save time? It would be great if anyone could share their ideas and help me out.

$\begingroup$I’ld like to point out that your first and second sentences contradict themselves. Take for example $\ce{SF4}$. We have two 2-electron-2-centre (2e2c) $\ce{S-F}$ $\sigma$ bonds and one (also 4e3c $\ce{F\bond{...}S\bond{...}F}$ bond. We also have one additional lone pair. The 4e3c bond is also $\sigma$-symmetric. Therefore, we have three or four $\sigma$ bonds — depending on how you count — and thus a steric number of four or five. However, sulphur is $\mathrm{sp^2}$ hybridised, i.e. only three orbitals take part in hybridisation.$\endgroup$
– JanJun 18 '17 at 17:04

$\begingroup$This question was posted before I had learned about the concept of banana bonds, and other special bonds, in which multiple centres are present (such as 4e3c, and 2e3c). Hence, I assumed that in all compounds, steric number equals the number of hybridised orbitals.$\endgroup$
– AbhigyanCJun 19 '17 at 11:54

3 Answers
3

The steric number is a property of an atom, not a compound. You need to know what an atom connected to a given atom to know its steric number. For simple compounds, you can usually determine these connections because the formula suggests a central atom and surrounding groups. For hydrocarbons and other organic compounds, you need to consider isomerism. Given the capability of carbon to form complicated bonding patterns, even simple formulas can produce a fair number of isomers with different bonding patterns and steric numbers.

Let's look at some examples.

$\ce{C4H10}$

This formula corresponds to two compounds with the structures shown:

In this case, both compounds have all four carbon atoms with steric number of 4.

it is not always true that a set of hydrocarbon isomers will always have the same steric number for all carbon atoms or even the same set of steric numbers.

$\ce{C4H8}$

This formula corresponds to six isomers:

Note that four of these structures have two carbon atoms with steric number 4 and two carbon atoms with steric number 3. The other two have all four carbon atoms with steric number 4.

Any method to calculate steric number for carbon atoms in an organic compound using just the formula will fail. You must examine the structure.

All right … I found myself a shortcut, and would like to share this in case it is useful for others. However, this formula is applicable to molecules with only one central atom.

Here is how it goes:

Find $N=\frac{V+M \pm I}{2}$, where $V = n(\ce{e-})$, the number of valence electrons of central atom, which is equal to the group number according to the old IUPAC system, $M = n(\text{atom})$, the number of monovalent atoms directly bonded to it, and $I$ is the number of positive or negative charges present (subtract it if the charge is positive, and add it if the charge is negative). This $N$ is the Steric Number.

Now, find the number of Bond Pairs ($BP$) of electrons, which is equal to the number of atoms surrounding the central atom. However, this is a little difficult for a species like $\ce{H3BO3}$, which is actually $\ce{B(OH)3}$, when written according the IUPAC method of writing the less electronegative atoms first.

Next, find the number of Lone Pairs ($LP$) of electrons, which is equal to $N-BP$.

Now, draw the structure of the atom, using the central atom, drawing the skeleton of the atom using the steric number, and then assigning the Bond Pairs and Lone pairs to the respective bonds/atoms.

That's for an atom with a single central atom.

Now, for a Hydrocarbon, albeit it is not possible to get the shape directly from the molecular formula, it is possible to find its structure and hybridisation if and only if the basic structure of the atom is provided.

For a compound with a single $\sigma$ bond between Carbon atoms, the hybridisation is $sp^3$

For one $\sigma$ and one $\pi$ bond, it is $sp^2$ hybridised, and

For one $\sigma$ and two $\pi$ bonds, it is $sp$ hybridised.

So, essentially, there is no formula for hydrocarbons, but there is a formula for smaller compounds, with a single central atom only.